Learning how summations work Before, I begin I want say that the original question asks, "A couple decides to continue to have children until a daughter is born. What is the expected number of children until a daughter is born?"
Below, is the solution that they had in the back of the textbook (Statistical Inference, Second Edition, Roger L. Berger, George Casella)

What I do not understand is how to deduce summations that well.
First of all, how do they get
$$\sum^\infty_{k=1}k(1-p )^{k-1}p = p - \sum^\infty_{k=1}\frac{d}{dp}(1-p )^{k}$$
Next, How do they simplify this as well $$-p\frac{d}{dp}{\bigg[\sum^\infty_{k=0}(1-p)^k-1\bigg]} = -p\frac{d}{dp}\bigg[\frac{1}{p}-1\bigg] = \frac{1}{p}$$
I would like this to be shown and explained intuitively because the statistics course I am taking requires me to be able to deduce summations on my own, and it seems to be a huge part of understanding the subject. Thank you ahead of time.
 A: As an alternate method that does not require the use of Geometric Series:
Let $E$ denote the answer, then consider the possible outcomes from the first child.  Either the child is a daughter ($p$) in which case the answer is $1$ or it is a son ($1-p$) and the answer is now $1+E$.
Thus $$E=p\times 1 + (1-p)\times (E+1)\implies E=\frac 1p$$
In general, the use of Geometric Series only makes sense in those situations (as here) in which one can easily enumerate all the possible outcomes.  Generally speaking, however, this is not possible.
A: This is a long-winded comment that is intended as a supplement to the other already posted responses.  This is not an answer.
The foundation of the geometric series is the idea that for any real number $x \neq 1$, and any positive integer $n$, you have that
$$\left(1 + x + x^2 + \cdots + x^n\right) \times (1 - x) = \left[1 - x^{(n+1)}\right]. \tag1 $$
Formulas around geometric series are all based on equation (1), above.
For example, from equation (1), you have that
$$\left(1 + x + x^2 + \cdots + x^n\right) = \frac{1 - x^{(n+1)}}{1 - x}. \tag2 $$
Now, for $~x \neq 1, ~n \in \Bbb{Z^+},~$ let $E_n(x)$ denote
$\displaystyle \frac{x^{(n+1)}}{1 - x}$.
Then you have that
$$\left(1 + x + x^2 + \cdots + x^n\right) = \frac{1}{1 - x} - E_n(x). \tag3 $$
This means that when $\displaystyle \left(1 + x + x^2 + \cdots + x^n\right)$
is estimated as $\displaystyle \frac{1}{1-x}$,
then $E_n(x)$ is the error term in the estimation.
Therefore, a natural question to ask is : 
as $n \to \infty$, for which values of $x$ will $E_n(x)$ go to $(0)$?  
[Denote this question as Q1].
Note that since fixed values of $x \neq 1$ are being considered, question Q1 is equivalent to asking:
as $n \to \infty$, for which values of $x$ will $x^{(n+1)}$ go to $(0)$? 
[Denote this question as Q2].
It is an intermediate result in Real Analysis (AKA Calculus) that the answer to question Q2 is that 
as $n \to \infty, ~x^{(n+1)} \to (0) \iff |x| < 1.$
What this implies is that for any $|x| < 1,$ the error term in equation (3) above will go to $(0)$ as $n \to \infty.$
This implies that for all $|x| < 1,$ you have that
$\displaystyle \lim_{n \to \infty} \sum_{k=0}^n x^k = \frac{1}{1-x}.$
A: 
First of all, how do they get
$$\sum^\infty_{k=1}k(1-p )^{k-1}p = p - \sum^\infty_{k=1}\frac{d}{dp}(1-p )^{k}$$

By way of a typo.  That is not what they should get.
$$\begin{align}\sum^\infty_{k=1}k(1-p )^{k-1}p ~&=~ p\sum_{k=1}^\infty k(1-p)^{k-1}\\ &=~-p\sum_{k=1}^\infty \dfrac{\mathrm d (1-p)^k}{\mathrm d p}\\[3ex]&=~-p\dfrac{\mathrm d ~~}{\mathrm d p}\left(\sum_{k=1}^\infty(1-p)^k\right)\\&=~-p\dfrac{\mathrm d ~~}{\mathrm d p}\left(\sum_{k=0}^\infty(1-p)^k-(1-p)^0\right)\\&=~-p\dfrac{\mathrm d ~~}{\mathrm d p}\left(\dfrac 1p-1\right)\\&=~-p\left(-\dfrac 1{p^2}\right)\\&=\dfrac 1p \end{align}$$

Next, How do they simplify this as well $$-p\frac{d}{dp}{\bigg[\sum^\infty_{k=0}(1-p)^k-1\bigg]} = -p\frac{d}{dp}\bigg[\frac{1}{p}-1\bigg] = \frac{1}{p}$$

By way of change of variables.  When $\lvert 1-p\rvert \leqslant 1$ (which it is because $p$ is a probability measure), then the series converges and so we have:
$$\begin{align}\sum_{k=0}^\infty (1-p)^k~&=\sum_{j=0}^\infty (1-p)^j\\&=~(1-p)^0+\sum_{j=1}^\infty (1-p)^j\\&=~1+\sum_{k=0}^\infty (1-p)^{k+1}\\&=~1+(1-p)\sum_{k=0}^\infty (1-p)^k\\\therefore\quad(1-(1-p))\sum_{k=0}^\infty (1-p)^k~&=~1\\\sum_{k=0}^\infty(1-p)^k~&=~\dfrac 1p\end{align}$$
